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Not correct, they can choose to not accept any blocks containing banned transactions, and still have the longest chain (statistically speaking). That's why the

Not correct, they can choose to not accept any blocks containing banned transactions, and still have the longest chain (statistically speaking). That's why the 51% number is significant.

tboyd47 4y ago
That assumes the other 49% won't accept the bad guys' blocks in their chain. Those blocks are valid, so the 49% will accept them and build off of them. At any point there is still a 49% chance the next block will be mined by a good guy. If the bad guys decide not to accept the good guys' blocks, then they are hard forking Bitcoin and will end up just like Bitcoin Cash: irrelevant. This is the exact problem PoW was designed to solve, and it works very well, which is why a 51% attack has never succeeded or even been attempted against Bitcoin, and never will.
- proto-n 4y ago
  They are not hard-forking, it will be compatible with the original chain and original rules. There will be two competing forks, one censored, one not censored, and the censored one is longer (because 51%). Consensus rules say longer chain is valid. So it _is_ the official "bitcoin", according to consensus rules. The 51% attack is not some made up thing.
  - tboyd47 4y ago
    If 51% of miners decide to totally ignore the other 49% then their chain will most definitely not be the longest chain for any more than 30 minutes. Because the 49% are still using 100% of miners' work. The only way they can maintain that chain is by a hard fork.
    - gus_massa 4y ago
      It's easier to explain a 66% attract. The censored chain produce the double of blocks of the censored one. Each time the uncensored chain produce a block it's ignored by the censored one. So after a time, they are behind and have to jump to the other chain and ignore their own old blocks. Something like this graph: Censored blocks: - Uncensored Fork blocks: \ Uncensored Normal blocks: * Censored: *-------------------------- Uncensored: \** \* \*** \** \*
      - tboyd47 4y ago
        So, now that I've obliterated the 51% attack idea, shall we move on to 66% attack? It doesn't matter if it's a 90% attack. The minority miners are constantly receiving and using the majority's blocks. So it is not 66% vs. 33%, or 90% vs. 10%; it is 60% vs. 100% and 90% vs. 100%. All it takes is for the less-restrictive minority to mine two blocks in a row and they've permanently outpaced the majority's more restrictive chain.
        cesarb 4y ago
        > The minority miners are constantly receiving and using the majority's blocks. They can use the majority blocks at the cost of discarding their earlier blocks. Every time they accept a block from the majority, they undo all the work they had done earlier. > All it takes is for the less-restrictive minority to mine two blocks in a row and they've permanently outpaced the majority's more restrictive chain. All it takes then is for the majority to mine three blocks in a row. The probability is always on their side.
        3np 4y ago
        All it takes then is for the majority to mine three blocks in a row at the precise right moment, consistently enough that there is lower gas available in minority blocks than demand from sanctioned transactions. If that's what it comes down to (and there is reason to believe it won't), I imagine sanction addresses will still transact, just with higher latency and at higher cost than others.
        gus_massa 4y ago
        The blocks with the "sactioned" transactions will be ignored by most of the people, so they will only live in forks that everyone will ignore. Assuming enough hash rate and the exchanges (and the users) agree to ignore the "sactioned" transactions, then they will never be in the main chain, not even slowly.
        4y ago
    - proto-n 4y ago
      Let me make sure I get this right: if you have two chains, one is mined by 49% hash power, the other is mined by 51% hash power, you think the 51% one won't be the longer one?
      - tboyd47 4y ago
        As I've explained in nearly every comment, one is mined by 100% hashpower, the other is mined by 51% hashpower. The 49% are still using the other blocks, too.
        simiones 4y ago
        This is where your mistake is: they can't be using the other blocks as well, because they are erasing their own (uncensored) transactions by doing so. Let's say that MiningPoolA controls 51% of hashpower, and MiningPoolB controls 49%. MiningPoolA is refusing to mine transactions from/to some wallet W. Time T1: MiningPoolA: OldChain -- Block1(no W) MiningPoolB: OldChain -- Block1'(W receives 1BTC) still in progress Any client will accept the chain with Block1 (no transactions from/to W) Time T2 - if MiningPoolB tries to compete MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W) MiningPoolB: OldChain -- Block1'(W receives 1BTC) -- Block2' (parent=Block1') still in progress Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up Time T2 - if MiningPoolB decides to accept MiningPoolA's chain, but add the transaction in the second block: MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W) MiningPoolB: OldChain -- Block1(no W) -- Block2' (parent=Block1, adds transaction W receives 1BTC) still in progress Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up If a mining pool had 51+% of hashpower, they would always be mining the longest chain, no one would be able to compete with them and publish another block (in principle, at least; in practice, since mining is not entirely deterministic, someone else will occasionally win the lottery and propose a new block faster).
        lupire 4y ago
        It's a blockchain, not a blockbag.
      - 4y ago
  - 3np 4y ago
    > Consensus rules say longer chain is valid. I will not go into depth here but it's a commonly held view that this changes under PoS. Just flying by so hopefully sources are easily located
- remram 4y ago
  If the 49% mine a block, the 51% can ignore it and will eventually surpass the 49%'s chain's length long term, no matter how lucky they are short-term.
  - tboyd47 4y ago
    The 51% can ignore the 49%'s blocks, but there is no way the 51% can force the 49% to ignore THEIR blocks. The only way 51% of miners can fork the blockchain and actually overtake the other 49% with more restrictive rules is if they have a hidden nuclear reactor dedicated to mining Bitcoin that produces hashpower equal or more than 49% of the network and they flip it on at the moment they fork. Again, it doesn't matter how many people are convinced a 51% attack is possible on Bitcoin. It's never even been tried once. So the burden of proof is on you all. I'm just trying to help explain why this concept has never been proven.
    - DSMan195276 4y ago
      > The 51% can ignore the 49%'s blocks, but there is no way the 51% can force the 49% to ignore THEIR blocks. If the 51% can ignore the blocks from the 49%, then why does it matter what the 49% is doing? Occasionally the 49% will be able to mine a block and temporarily create the longest chain, but the 51% can always just ignore that block and continue mining off of the one before it - eventually their chain will be the longest and the block(s) from the 49% will be lost. Really this isn't theoretical, it happens all the time by accident when two blocks get mined off of the same previous block. When that happens one of the two blocks gets lost, and the only difference with this scenario is that the 51% have enough hash power to ensure that their blocks are the ones that always (eventually) win and the 49% blocks are always lost.
- samkon 4y ago
  Except that blocks are not “finalized”, like you’re imagining. With 51% of mining power, I can mine faster than the rest of the network, for an arbitrary number of blocks. The 51% attacker will have the longest chain for as long as they keep mining their chain because they’re chain will always (eventually) be the longest one.
  - tboyd47 4y ago
    51% < 100% The other 49% are still incorporating the 51%'s blocks in their work. There is no way 51% of miners can stay ahead of the combined hashpower of the entire Bitcoin mining population.
    - deleted 4y ago
      [deleted]
    - cesarb 4y ago
      > The other 49% are still incorporating the 51%'s blocks in their work. It's not that simple. That block from the 49% will be ignored by the 51% (since it contains "banned" transactions), which will continue the chain on the previous block (the "51% chain". Now the 49% has two options. If they continue to build the chain on top of that 49% block (the "49% chain"), after a while the 51% chain will be longer (because that side has the most hash power). The other option is to build again on top of the 51% chain, as you suggested (and AFAIK that's what unmodified Bitcoin software will do after a while); but to do that, they have to discard that block they had included earlier (since it's not in the 51% chain). That is: yes, the 49% can include "banned" transactions, but that inclusion will be undone later. They can include these transactions again, but that inclusion will be undone again. They can never get far enough for these blocks with the "banned" transactions to be permanent. > Again, a 51% attack has never even been attempted on the Bitcoin network despite huge potential monetary upside if it succeeds. First, this is not the "traditional" 51% attack, which involves mining an alternative longer chain in secret. Second, the most a 51% attack can do is double spend coins (or prevent them from being spent); converting that into real money requires spending the coin twice (for instance, sending coins to an exchange, withdrawing the resulting money, and then undoing the sending to the exchange so the attacker keeps the coins), and the monetary upside isn't that big in most scenarios. Third, the cost for doing that is not as small as you're thinking (start with the cost to obtain enough miners to have 51% of the hash power), which is why it hasn't AFAIK been attempted on Bitcoin (but AFAIK, it has been attempted on less popular networks which have small total hash power). And if you fail the attack, you have wasted all that cost.

Not correct, they can choose to not accept any blocks containing banned transactions, and still have the longest chain (statistically speaking). That's why the 51% number is significant.

tboyd47 4y ago

That assumes the other 49% won't accept the bad guys' blocks in their chain. Those blocks are valid, so the 49% will accept them and build off of them. At any point there is still a 49% chance the next block will be mined by a good guy. If the bad guys decide not to accept the good guys' blocks, then they are hard forking Bitcoin and will end up just like Bitcoin Cash: irrelevant. This is the exact problem PoW was designed to solve, and it works very well, which is why a 51% attack has never succeeded or even been attempted against Bitcoin, and never will.

proto-n 4y ago
They are not hard-forking, it will be compatible with the original chain and original rules. There will be two competing forks, one censored, one not censored, and the censored one is longer (because 51%). Consensus rules say longer chain is valid. So it _is_ the official "bitcoin", according to consensus rules. The 51% attack is not some made up thing.
- tboyd47 4y ago
  If 51% of miners decide to totally ignore the other 49% then their chain will most definitely not be the longest chain for any more than 30 minutes. Because the 49% are still using 100% of miners' work. The only way they can maintain that chain is by a hard fork.
  - gus_massa 4y ago
    It's easier to explain a 66% attract. The censored chain produce the double of blocks of the censored one. Each time the uncensored chain produce a block it's ignored by the censored one. So after a time, they are behind and have to jump to the other chain and ignore their own old blocks. Something like this graph: Censored blocks: - Uncensored Fork blocks: \ Uncensored Normal blocks: * Censored: *-------------------------- Uncensored: \** \* \*** \** \*
    - tboyd47 4y ago
      So, now that I've obliterated the 51% attack idea, shall we move on to 66% attack? It doesn't matter if it's a 90% attack. The minority miners are constantly receiving and using the majority's blocks. So it is not 66% vs. 33%, or 90% vs. 10%; it is 60% vs. 100% and 90% vs. 100%. All it takes is for the less-restrictive minority to mine two blocks in a row and they've permanently outpaced the majority's more restrictive chain.
      - cesarb 4y ago
        > The minority miners are constantly receiving and using the majority's blocks. They can use the majority blocks at the cost of discarding their earlier blocks. Every time they accept a block from the majority, they undo all the work they had done earlier. > All it takes is for the less-restrictive minority to mine two blocks in a row and they've permanently outpaced the majority's more restrictive chain. All it takes then is for the majority to mine three blocks in a row. The probability is always on their side.
        3np 4y ago
        All it takes then is for the majority to mine three blocks in a row at the precise right moment, consistently enough that there is lower gas available in minority blocks than demand from sanctioned transactions. If that's what it comes down to (and there is reason to believe it won't), I imagine sanction addresses will still transact, just with higher latency and at higher cost than others.
        gus_massa 4y ago
        The blocks with the "sactioned" transactions will be ignored by most of the people, so they will only live in forks that everyone will ignore. Assuming enough hash rate and the exchanges (and the users) agree to ignore the "sactioned" transactions, then they will never be in the main chain, not even slowly.
        4y ago
  - proto-n 4y ago
    Let me make sure I get this right: if you have two chains, one is mined by 49% hash power, the other is mined by 51% hash power, you think the 51% one won't be the longer one?
    - tboyd47 4y ago
      As I've explained in nearly every comment, one is mined by 100% hashpower, the other is mined by 51% hashpower. The 49% are still using the other blocks, too.
      - simiones 4y ago
        This is where your mistake is: they can't be using the other blocks as well, because they are erasing their own (uncensored) transactions by doing so. Let's say that MiningPoolA controls 51% of hashpower, and MiningPoolB controls 49%. MiningPoolA is refusing to mine transactions from/to some wallet W. Time T1: MiningPoolA: OldChain -- Block1(no W) MiningPoolB: OldChain -- Block1'(W receives 1BTC) still in progress Any client will accept the chain with Block1 (no transactions from/to W) Time T2 - if MiningPoolB tries to compete MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W) MiningPoolB: OldChain -- Block1'(W receives 1BTC) -- Block2' (parent=Block1') still in progress Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up Time T2 - if MiningPoolB decides to accept MiningPoolA's chain, but add the transaction in the second block: MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W) MiningPoolB: OldChain -- Block1(no W) -- Block2' (parent=Block1, adds transaction W receives 1BTC) still in progress Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up If a mining pool had 51+% of hashpower, they would always be mining the longest chain, no one would be able to compete with them and publish another block (in principle, at least; in practice, since mining is not entirely deterministic, someone else will occasionally win the lottery and propose a new block faster).
      - lupire 4y ago
        It's a blockchain, not a blockbag.
    - 4y ago
- 3np 4y ago
  > Consensus rules say longer chain is valid. I will not go into depth here but it's a commonly held view that this changes under PoS. Just flying by so hopefully sources are easily located
remram 4y ago
If the 49% mine a block, the 51% can ignore it and will eventually surpass the 49%'s chain's length long term, no matter how lucky they are short-term.
- tboyd47 4y ago
  The 51% can ignore the 49%'s blocks, but there is no way the 51% can force the 49% to ignore THEIR blocks. The only way 51% of miners can fork the blockchain and actually overtake the other 49% with more restrictive rules is if they have a hidden nuclear reactor dedicated to mining Bitcoin that produces hashpower equal or more than 49% of the network and they flip it on at the moment they fork. Again, it doesn't matter how many people are convinced a 51% attack is possible on Bitcoin. It's never even been tried once. So the burden of proof is on you all. I'm just trying to help explain why this concept has never been proven.
  - DSMan195276 4y ago
    > The 51% can ignore the 49%'s blocks, but there is no way the 51% can force the 49% to ignore THEIR blocks. If the 51% can ignore the blocks from the 49%, then why does it matter what the 49% is doing? Occasionally the 49% will be able to mine a block and temporarily create the longest chain, but the 51% can always just ignore that block and continue mining off of the one before it - eventually their chain will be the longest and the block(s) from the 49% will be lost. Really this isn't theoretical, it happens all the time by accident when two blocks get mined off of the same previous block. When that happens one of the two blocks gets lost, and the only difference with this scenario is that the 51% have enough hash power to ensure that their blocks are the ones that always (eventually) win and the 49% blocks are always lost.
samkon 4y ago
Except that blocks are not “finalized”, like you’re imagining. With 51% of mining power, I can mine faster than the rest of the network, for an arbitrary number of blocks. The 51% attacker will have the longest chain for as long as they keep mining their chain because they’re chain will always (eventually) be the longest one.
- tboyd47 4y ago
  51% < 100% The other 49% are still incorporating the 51%'s blocks in their work. There is no way 51% of miners can stay ahead of the combined hashpower of the entire Bitcoin mining population.
  - deleted 4y ago
    [deleted]
  - cesarb 4y ago
    > The other 49% are still incorporating the 51%'s blocks in their work. It's not that simple. That block from the 49% will be ignored by the 51% (since it contains "banned" transactions), which will continue the chain on the previous block (the "51% chain". Now the 49% has two options. If they continue to build the chain on top of that 49% block (the "49% chain"), after a while the 51% chain will be longer (because that side has the most hash power). The other option is to build again on top of the 51% chain, as you suggested (and AFAIK that's what unmodified Bitcoin software will do after a while); but to do that, they have to discard that block they had included earlier (since it's not in the 51% chain). That is: yes, the 49% can include "banned" transactions, but that inclusion will be undone later. They can include these transactions again, but that inclusion will be undone again. They can never get far enough for these blocks with the "banned" transactions to be permanent. > Again, a 51% attack has never even been attempted on the Bitcoin network despite huge potential monetary upside if it succeeds. First, this is not the "traditional" 51% attack, which involves mining an alternative longer chain in secret. Second, the most a 51% attack can do is double spend coins (or prevent them from being spent); converting that into real money requires spending the coin twice (for instance, sending coins to an exchange, withdrawing the resulting money, and then undoing the sending to the exchange so the attacker keeps the coins), and the monetary upside isn't that big in most scenarios. Third, the cost for doing that is not as small as you're thinking (start with the cost to obtain enough miners to have 51% of the hash power), which is why it hasn't AFAIK been attempted on Bitcoin (but AFAIK, it has been attempted on less popular networks which have small total hash power). And if you fail the attack, you have wasted all that cost.